Sunday, September 27, 2009

99% sure??

When someone says they are 99% sure of something, they probably don't mean that. If they were 99% sure, then they would be willing to bet on it with 99-to-1 odds. That is, they'd be willing to bet $99 for the chance to win $1.

First, let's be clear that you always get back your original bet plus ($99) what you win ($1). When I say "I'll give you x-to-y odds" that means "If I win, you give me y, and if you win I give you x".

It's all about the expected value. Let's say you bet x ($99) for a chance to win y ($1), and the probability of winning is p (0.99). If you lose you will be x poorer, and if you win you will y richer. The expected value is

E = p*y - (1 - p)*x

In this case, that is 0 (break even).

When you say you're uncomfortable risking more money than you'll get in return, then it just sounds like you're not confident in your calculation of p, so let's do an example in which you would be confident in p: dice rolling.

Roll a die. If you roll higher than a 1, then I will give you $1. If you roll a 1, you have to give me $4. So, p = 5/6, x = $4, y = $1, and thus E = 5/6 * 4 - 1/6 * 1 = 1/6.

So, there's a case where I'm giving you worse than 50-50 odds (1-to-4), but you'd be a fool not to engange me in that bet!

Similarly, if I gave you 10-to-1 odds and you have only a 1% chance of winning, you wouldn't want to engange in that bet.

Pot odds are calculated the exact same way. Here are a couple things that you gotta watch out for:

1. Anything you and your opponent put in the pot in the previous round of betting is part of y, not x! Once you have put money in the pot, it is considered *not yours*, but *winnable*.

2. If there are more rounds of betting after the current one, you have to factor those in, and that gets complicated.